Set Operations with Empty sets I have a question regarding performing set operations on empty sets.
For example let A = ∅, Let B = {A, ∅}, Let C = {A, B}.
Would B = {∅, ∅} and C = {∅, {∅,∅}}? or would B = {∅} and C = {∅, {∅}} since {∅, ∅} reduces to {∅}?
Now if I wanted to do $A \cup C$ would the answer be $∅ \cup C$ -> C = {∅, {∅}}?
Now what if I wanted to do D = C - B, would the result be {{∅}}?
One last question. If P = {∅}, Q = {P}, R = {∅,P}, S = {∅, P, Q, R};
What would X = {$ x : (x ∈ R) \wedge ( x ⊆ S)$}?
My thoughts are that Q = {{∅}}, R = {∅, {{∅}}} and S = { ∅, {∅}, {{∅}}, {∅, {{∅}}} }


*

*x ∈ R would mean x = ∅ and x = {{∅}}

*x ⊆ S would mean x is a set containing any combination of ∅, {∅},
{{∅}}, {∅, {{∅}}} as members


Therefore X = {{{∅}}} as {{∅}} is the only member that fits both those conditions?
I've tried looking at some lectures online on set theory but I want to make sure my understanding is correct so far.
On a more general note, I am also wondering if lets say D = {1, 2} and E = {D, 5}. Would E = {1,2,5} or E = {{1,2},5}?
Thanks
 A: $B$ contains a single element which is an empty set (or, equivalently, is equal to $A$): $B = \{\emptyset\} = \{A\}$
$C$ contains two elements, one of which is an empty set (or, equivalently, is equal to $A$), and another is $B$: $C = \{\emptyset, \{\emptyset\}\}$.
$A \cup C$ contains all elements $A$ contains (there are no such elements, as $A$ is empty) and all elements $C$ contains: $A \cup C = C = \{\emptyset, \{\emptyset\}\}$.
$C = \{A\} \sqcup \{B\}$, and $B = \{A\}$, hence $C - B = (\{A\} \sqcup \{B\}) - \{A\} = \{B\} = \{\{A\}\} = \{\{\emptyset\}\}$.
As for the last question, $x \in R$ is equal to $x = \emptyset \vee x = P$ per $R$ definition. Empty set is obviously a subset of any set (including $S$); and $P \subseteq S$ because all elements $P$ contains (there is only one of them - an empty set) are also elements of $S$. So $X = R$.

On a more general note, I am also wondering if lets say D = {1, 2} and E = {D, 5}. Would E = {1,2,5} or E = {{1,2},5}?

$E$ is a two-element set (as long as $D$ is not equal to $5$): one element is $5$ and another is $D$. Thus, $E = \{\{1, 2\}, 5\}$.
